Question

When only two treatments are involved, ANOVA and the Student’s t test (Chapter 11) result in the same conclusions. Also, for computed test statistics, t^₂ = F. To demonstrate this relationship, use the following example. Fourteen randomly selected students enrolled in a history course were divided into two groups, one consisting of 6 students who took the course in the normal lecture format. The other group of 8 students took the course as a distance course format. At the end of the course, each group was examined with a 50-item test. The following is a list of the number correct for each of the two groups.

Traditional Lecture	Distance
35	47
32	31
37	48
31	30
30	47
38	33
	42
	41

  

   

1. a-1. Complete the ANOVA table. (Round your SS, MS, and F values to 2 decimal places and p value to 4 decimal places.)

	Source SS df MS F p Factors Error Total

1. a-2. Use a α = 0.01 level of significance. (Round your answer to 2 decimal places.)

	The critical value of F___

2. Using the t test from Chapter 11, compute t. (Negative amount should be indicated by a minus sign. Round your answer to 3 decimal places.)

3. There is any difference in the mean test scores.

	(fail to reject, reject) H0. There is (no statistically significant difference, a statistically significant difference) in the mean scores between lecture and internet-based formats.(

Answer 1

a-1_) applying one way ANOVA:

Source of Variation	SS	df	MS	F	P-value
Between Groups	125.15	1	125.15	3.32	0.0932
Within Groups	451.71	12	37.64
Total	576.86	13

a-2()

The critical value of F =9.33

b)

t =-1.823

c)

fail to reject , there is no statistically significant difference in the mean scores between lecture and internet-based formats